Answer:
The magnetic field is strongest at the center and weakest between the two poles just outside the bar magnet. The magnetic field lines are densest at the center and least dense between the two poles just outside the bar magnet.
Explanation:
<span> d = r*t is the basic distance equation
d = 6000 km
t with the tail wind = 6 hr
r with the tail wind = speed of the plane + wind speed = s + w
t with the head wind = 7.5 hr
r with the head wind = speed of the plane - wind speed = s-w
(s+w)*6 = 6000
(s-w)*7.5 = 6000
s + w = 1000
s - w = 800
</span><span> 2s = 1800
s = 900 km/h
s + w = 1000
w = 100
Check the anwer by calculating the return trip.
(900-100) * 7.5 = 800 * 7.5
800 * 7.5 = 6000 km
Answer: The rate of the jet in still air is 900 km/h. The rate of the wind is 100 km/hr.</span>
Good morning.
We have:

Where
j is the unitary vector in the direction of the
y-axis.
We have that

We add the vector
-a to both sides:

Therefore, the magnitude of
b is
47 units.
Answer:
charge on each
Q1 = 2.06 ×
C
Q2 = 7.23 ×
C
when force were attractive
Q1 = 1.07 ×
C
Q2 = -1.39 ×
C
Explanation:
given data
total charge = 93.0 μC
apart distance r = 1.14 m
force exerted F = 10.3 N
to find out
What is the charge on each and What if the force were attractive
solution
we know that force is repulsive mean both sphere have same charge
so total charge on two non conducting sphere is
Q1 + Q2 = 93.0 μC = 93 ×
C
and
According to Coulomb's law force between two sphere is
Force F =
.........1
Q1Q2 = 
here F is force and r is apart distance and k is 9 ×
N-m²/C² put all value we get
Q1Q2 = 
Q1Q2 = 1.49 ×
C²
and
we have Q2 = 93 ×
C - Q1
put here value
Q1² - 93 ×
Q1 + 1.49 ×
= 0
solve we get
Q1 = 2.06 ×
C
and
Q1Q2 = 1.49 ×
2.06 ×
Q2 = 1.49 ×
Q2 = 7.23 ×
C
and
if force is attractive we get here
Q1Q2 = - 1.49 ×
C²
then
Q1² - 93 ×
Q1 - 1.49 ×
= 0
we get here
Q1 = 1.07 ×
C
and
Q1Q2 = - 1.49 ×
2.06 ×
Q2 = - 1.49 × 
Q2 = -1.39 ×
C